Question

Find the value of k for which each of the following systems of equations have infinitely many solutions
kx - 2y + 6 = 0 and 4x - 3y + 9 =0​

Answer 1

GIVEN :

Two Equations :

1) kx - 2y + 6 = 0

2) 4x - 3y + 9 =0

It is said that the equations have infinitely many solutions.

Then,

$\frac{a1}{a2} = \frac{b1}{a2} = \frac{c1}{c2} \\ \\ a1 = k \: \: \: \: b1 = - 2 \: \: \: \: c1 = 6 \\ \\ a2 = 4 \: \: \: b2 = - 3 \: \: \: c2 = 9$

$\frac{k}{4} = \frac{ - 2}{ - 3}$

Cross Multiply the above terms.

$\frac{k}{4} = \frac{ - 2}{ - 3} \\ \\ k \times - 3 = - 8 \\ \\ - 3k = - 8 \\ \\ k = \frac{ - 8}{ - 3} \\ \\ k = \frac{8}{3}$

Therefore, the value of k is 8/3.

Answer 2

Name of the chapter - "Pair of Linear Equations in two variables"

Class - 10

$\underline{\underline{\bf SOLUTION :}}$

A pair of linear equations are having infinitely many solutions if

$\bf \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

So

given :

Two pair of linear equations are

1. kx - 2y + 6 = 0

2. 4x - 3y + 9 = 0

a₁ = k,  b₁ = -2, c₁ = 6

a₂ = 4, b₂ = -3, c₂ = 9

$\bf \frac{k}{4}=\frac{6}{9}$

$\bf \frac{k}{4} = \frac{2}{3}$

By doing cross multiplication we get

3k = 2×4

3k = 8

$\bf k=\frac{8}{3}$

Hence,

The value of k should be 8/3 to get infinity number of solutions for the given pair of linear pairs.